So the number of equivalence classes is the same as the number of integers, am I right? Because that's how many differet "floors" we have. Here I have another relation defined on the set of these classes, meaning

And a few questions to answer.

I. Prove that the definition of the relation is correct.
This means I have to prove that no matter which member of the equivalence class I choose, the relation between equivalence classes will stand, meaning that for if then for any are too in relation .
We know that and . So there's two cases to analyze, when two equivalence classes are related by the first part of the logical disjunction, then , so the definiton is correct. For the second part of the logical disjunction we start by stating that , because otherwise would be in the equivalence class of and fulfill the first part of the disjunction, any by extensions, whic means , so the definition is correct in this case also.

II. Prove that this relation is a linear order, meaning that for any two equivalence classes either or . It suffices to say that if two numbers have the same floor then their equivalence classes are in relation, if not than one must be larger than the other so it would still fulfill the definition.

III. Show, that is isomorphic with , which is an order of integers by normal greater-equal relation. Well the bijection in this case to a equivalence class assigns an integer .